Question

G(x) = -2x^3 – 15x^2 + 36x

Answer 1

Consider the function $G(x) = -2x^3 - 15x^2 + 36x.$ First, factor it:

$G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).$

The x-intercepts are at points $\left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).$

1. From the attached graph you can see that

• function is positive for $x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);$
• function is negative for $x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).$

2. Since

$G(-x) = -2(-x)^3 - 15(-x)^2 + 36(-x)=2x^3-15x^2-36x\neq G(x)\ \text{and }\neq -G(x)$ the function is neither even nor odd.

3. The domain is $x\in (-\infty,\infty),$ the range is $y\in (-\infty,\infty).$

Answer 2

a) G(x) factors as x(-2x^2 -15x+36) so has a zero at x=0 and at values of x revealed by the quadratic formula:

For ax²+bx+c=0, the solutions are

... x = (-b±√(b²-4ac))/(2a)

Here, we have a=-2, b=-15, c=36, so

... x = (15±√(225+288))/(-4) = -3.75±√32.0625

... x ≈ {-9.412, 1.912}

The function is positive for x in (-∞, -9.412) ∪ (0, 1.912).

The function is negative for x in (-9.412, 0) ∪ (1.912, ∞).

b) The function contains both even-degree and odd-degree terms, so has no even or odd symmetry. (A cubic always has odd symmetry about its point of inflection, but that point is not x=0 for this function.)

c) The domain and range of any odd-degree polynomial are all real numbers.